The Cartesian equation of the curve x=3+5 cos | vedclass

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(D) Given equations are $x=3+5 \cos 	heta$ and $y=2+5 \sin 	heta$. Rearranging the terms,we get $\frac{x-3}{5} = \cos 	heta$ and $\frac{y-2}{5} = \

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$x^{2}+y^{2}-6x-4y-12=0$

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(D) Given equations are $x=3+5 \cos 	heta$ and $y=2+5 \sin 	heta$. Rearranging the terms,we get $\frac{x-3}{5} = \cos 	heta$ and $\frac{y-2}{5} = \sin 	heta$. Using the identity $\cos^{2} 	heta + \sin^{2} 	heta = 1$,we substitute the expressions: $(\frac{x-3}{5})^{2} + (\frac{y-2}{5})^{2} = 1$. Expanding the squares: $\frac{x^{2}-6x+9}{25} + \frac{y^{2}-4y+4}{25} = 1$. Multiplying by $25$: $x^{2}-6x+9 + y^{2}-4y+4 = 25$. Simplifying: $x^{2}+y^{2}-6x-4y+13 = 25$. Thus,the Cartesian equation is $x^{2}+y^{2}-6x-4y-12 = 0$.

The Cartesian equation of the curve $x=3+5 \cos \theta, y=2+5 \sin \theta$ is $(0 \leq \theta \leq 2 \pi)$.

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